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olasank [31]
3 years ago
7

Pls Help! John wants to prove ABC = to DEF. He Knoes AB = DE and AC = DF. With Screenshot

Mathematics
1 answer:
matrenka [14]3 years ago
7 0

Answer:

\angle A \cong \angle D

Step-by-step explanation:

The Side-Angle-Side method cana only be used when information given shows that an included angle which is between two sides of a ∆, as well as the two sides of the ∆ are congruent to the included side and two sides of the other ∆.

Thus, since John already knows that \overline{AB} \cong \overline{DE} and \overline{AC} \cong \overline{DF}, therefore, an additional information showing that the angle between \overline{AB} and \overline{DE} in ∆ABC is congruent to the angle between \overline{AC} and \overline{DF} in ∆DEF.

For John to prove that ∆ABC is congruent to ∆DEF using the Side-Angle-Side method, the additional information needed would be \angle A \cong \angle D.

See attachment for the diagram that has been drawn with the necessary information needed for John to prove that ∆ABC is congruent to ∆DEF.

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Answer:

all work is shown and pictured

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3 years ago
Any body can help me with this ?
ValentinkaMS [17]
Do 12x-9 + 141 = 180 (add -9 and 4)
12x +132= 180 (Subtract 132 from 180)
12x =48 (Divide 48 by 12)
x=4
6 0
3 years ago
Jessica plans to purchase a car in one year at a cost of $30,000. how much should be invested in an account paying 10% compounde
dezoksy [38]
The formula is
A=p (1+r/k)^kt
A fund needed 30000
p Amount invested?
R interest rate 0.1
K compounded semiannual 2
T time 1 year
Solve the formula for p to get
P=A÷(1+r/k)^kt
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4 0
3 years ago
Find the zeros of y=x^2-8x-3​
Anton [14]
This is gonna be the answer. It’s 4+sqrt(19) and 4-sqrt(19). Which is approximately 8.36 and -0.36. In x intercept form, it’s (8.36,0) and (-0.36,0) also I solved this by completing the square. I hope this helps!!! Mark as brainliest!!!

4 0
3 years ago
Suppose a certain population satisfies the logistic equation given by dP
Ksenya-84 [330]

Answer:

The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

\frac{dP}{dt}=10P-P^2

with P(0)=1. We need to find the population when t=3.

Using variable separable method we get

\frac{dP}{10P-P^2}=dt

Integrate both sides.

\int \frac{dP}{10P-P^2}=\int dt             .... (1)

Using partial fraction

\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}

A=\frac{1}{10},B=\frac{1}{10}

Using these values the equation (1) can be written as

\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt

\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt

On simplification we get

\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C

\frac{1}{10}(\ln \frac{P}{10-P})=t+C

We have P(0)=1

Substitute t=0 and P=1 in above equation.

\frac{1}{10}(\ln \frac{1}{10-1})=0+C

\frac{1}{10}(\ln \frac{1}{9})=C

The required equation is

\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})

Multiply both sides by 10.

\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}

e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}

\frac{P}{10-P}=\frac{1}{9}e^{10t}

Reciprocal it

\dfrac{10-P}{P}=9e^{-10t}

P(t)=\dfrac{10}{1+9e^{-10t}}

The population when t = 3 is

P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}

Using calculator,

P=9.999\approx 10

Therefore, the population when t = 3 is 10.

8 0
3 years ago
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